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Tetrahedral numbers (https://oeis.org/A000292) are:

1, 4, 10, 20, 35, 56, 84, etc

WHERE:

$$mathrmTetrahedral(n) = frac(n(n+1)(n+2))6$$

But what is the test for whether a given number is a tetrahedral number?
I also need to know, if it is a tetrahedral number, then which term in the tetrahedral sequence is it? I am looking for a formula or something that would enable me to get these results:

input result ("is tetrahedral")
1 1
2 false
3 false
4 2
5 false
6 false
7 false
8 false
9 false
10 3
11 false
12 false

I can't find any formula on the Internet for this. Any pointers in the right direction would be great.

Another way to look at that: $fracn(n+1)(n+2)6=frac(n+1)^3-(n+1)6$.

So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]6mrceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.

Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]6krfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.

This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:

1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]6krfloor$.

2) If $textTetra(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $textTetra(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.

3) If $textTetra(n_0)<k$, begin testing all naturals $n>n_0$ until $textTetra(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.

4) If $textTetra(n_0)=k$, then you are done, and the index is $n_0$.

For a given $k$, you want to know if it exists an integer $n$ such that
$$fracn(n+1)(n+2)6=k$$ A simple way is to look at the cubic equation
$$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
$$n=-1+frac2 sqrt3cosh left(frac13 cosh ^-1left(9 sqrt3,
kright)right)$$ If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.

If you prefer "simpler" analytical formulae, you also have
$$n=-1+t_1+t_2$$ where
$$t_1^3=frac13 left(sqrt3 sqrt243 k^2-1+27 kright)qquad textand qquad t_2^3=frac19 left(sqrt3 sqrt243 k^2-1+27 kright)$$

Assume that $n(n+1)(n+2)/6=k$. Then $n(n+1)(n+2)-6k=0$, which has exactly one positive root for any positive $k$ (the left side increases monotonically and without bound from a negative intercept for positive $n$). If this is to be an integer then the rational root theorem guarantees that it will be a divisor of $6k$. So test such divisors in increasing order until you hit $n(n+1)(n+2)=6k$ or overshoot with $n(n+1)(n+2)>6k$.

We can cut down on trials by using the property that a tetrahedral number us the sum of alternating squares. With an odd argument the sum is $1^2+3^2+5^2+...$ with the sum truncated at the argument, thus the fifth nonzero terrahedral number is $1^2+3^2+5^2$. With an even argument we would use $2^2+4^2+6^2+...$.

If the argument $n$ is odd then, because all odd squares are $equiv 1bmod 8$, we must have $nequiv 2k-1bmod 8$. An even $n$ gives squares alternating between $4$ and $0bmod 8$, so if $k$ is divisible by $8$ then $nin0,6bmod 8$. If $k$ is instead four times an odd number then then $nin2,4bmod 8$.

Let's apply these ideas to $84$. This is four times an odd number, so an odd argument must be $equiv 7bmod 8$ and an even argument must be $in2,4bmod 8$. The divisors of $6×84=504$ satisfying these modular requirements are

$2,4,7,12,18,...,252$

We begin testing these in ascending order. We might see that $(8)(8+1)(8+2)>8^3>504$ (the cube root test in another answer), so either $2,4$ or $7$ works or else nothing does.